微积分 BC · 第5单元：微分的分析应用 · 阅读约 14 分钟 · 更新于 2026-05-11

一阶导数判别法判断相对极值 — AP 微积分 BC

AP 微积分 BC · 第5单元：微分的分析应用 · 14 min read

1. 一阶导数判别法的核心概念 ★★☆☆☆ ⏱ 3 min

一阶导数判别法（也称为局部/相对极值判别法）是AP微积分BC第5单元的核心方法，在选择题和自由作答题中都会出现，占考试总分的4-7%。

该方法利用一阶导数的几何意义：$f'(x)$告诉我们$f(x)$是递增（$f'(x) > 0$）还是递减（$f'(x) < 0$）。通过分析临界点周围$f'(x)$的符号变化，我们可以将该点分类为相对极大值、相对极小值或两者都不是。

Exam tip: 在AP自由作答题中，你必须明确提及一阶导数的符号变化，才能获得极值分类的全部证明分数。

2. 内部临界点的符号分析 ★★☆☆☆ ⏱ 4 min

要对内部临界点应用一阶导数判别法，将$f'$的定义域按临界点分割为多个区间，在每个区间内取一点测试$f'(x)$的符号，然后使用以下符号变化规则进行分类：

如果$f'$在$c$处由正变负，则$f$在$x=c$处有一个**相对极大值**
如果$f'$在$c$处由负变正，则$f$在$x=c$处有一个**相对极小值**
如果$f'$在$c$处符号不变，则$f$在$x=c$处**没有相对极值**

Exam tip: 仅说明$f'(c)=0$不足以证明自由作答题中的极值存在。

3. $f'$不存在处的临界点 ★★★☆☆ ⏱ 4 min

只要$c$在$f$的定义域中，即使$f'(c)$不存在，也可能是临界点。一阶导数判别法对这些点的处理流程与$f'(c)=0$的点完全相同。这种情况常见于绝对值函数、根式函数和分段函数，在AP考试中经常出现。

📐 Worked Example

Let $f(x) = x^{2/3}(x - 4)$. Find all critical points and classify each using the First Derivative Test.

使用乘积法则计算$f'(x)$：
$f'(x) = \frac{2}{3}x^{-1/3}(x - 4) + x^{2/3} = \frac{5x - 8}{3x^{1/3}}$
识别临界点：$f'(x)$在$x=0$处不存在，且$f(0)=0$（因此$x=0$在$f$的定义域中），所以它是一个有效的临界点。令分子等于零得到$x=8/5$，此处$f'=0$且$f$有定义，因此$x=8/5$也是一个临界点。
对区间$(-\infty, 0)$, $(0, 8/5)$, $(8/5, \infty)$进行符号分析：
- At $x=-1$: $f'(-1) = \frac{-13}{-3} > 0$<br>- At $x=1$: $f'(1) = \frac{-3}{3} < 0$<br>- At $x=2$: $f'(2) = \frac{2}{3 \cdot 2^{1/3}} > 0$
分类：在$x=0$处，$f'$由正变负 → $(0, 0)$处为相对极大值。在$x=8/5$处，$f'$由负变正 → $(1.6, \approx -3.28)$处为相对极小值。

Exam tip: 在将$f'$不存在的点称为临界点之前，一定要确认该点在$f$的定义域中。

4. 端点极值的一阶导数判别法 ★★★☆☆ ⏱ 4 min

在闭区间$[a, b]$上，端点$x=a$和$x=b$可以是相对极值，因为定义相对极值时我们只考虑区间内的函数值。一阶导数判别法可以自然推广到端点，只需检查端点内侧$f'$的符号即可：

Left endpoint $x=a$: If $f'(x) < 0$ just right of $a$, $f(a)$ is a relative maximum. If $f'(x) > 0$ just right of $a$, $f(a)$ is a relative minimum.
Right endpoint $x=b$: If $f'(x) > 0$ just left of $b$, $f(b)$ is a relative maximum. If $f'(x) < 0$ just left of $b$, $f(b)$ is a relative minimum.

📐 Worked Example

Find all relative extrema of $f(x) = 2x^3 - 3x^2$ on the closed interval $[-1, 2]$, including endpoints.

Find interior critical points: $f'(x) = 6x^2 - 6x = 6x(x-1)$, so interior critical points at $x=0$ and $x=1$.
Sign analysis for interior points: Intervals $(-1, 0)$, $(0, 1)$, $(1, 2)$. Testing gives $f'(-0.5) > 0$, $f'(0.5) < 0$, $f'(1.5) > 0$. So $x=0$ is a relative maximum, $x=1$ is a relative minimum.
Classify endpoints: Left endpoint $x=-1$ has $f'(x)$ positive just right of $-1$, so $f$ increases away from $x=-1$ → relative minimum at $(-1, -5)$. Right endpoint $x=2$ has $f'(x)$ positive just left of $2$, so $f$ increases towards $x=2$ → relative maximum at $(2, 4)$.
Final result: Relative minima at $(-1, -5)$ and $(1, -1)$; relative maxima at $(0, 0)$ and $(2, 4)$.

Exam tip: When asked to find all relative extrema on a closed interval, don't forget to classify endpoints—this is a common AP exam point trap.

5. AP风格的额外例题 ★★★★☆ ⏱ 6 min

📐 Worked Example

Let $f(x) = \frac{x^2}{x - 2}$. (a) Find all critical points of $f$. Justify your answer. (b) Classify each critical point as a relative maximum, relative minimum, or neither using the First Derivative Test. Justify your answer. (c) Identify all relative extrema of $f$ on the interval $[0, 4]$, including endpoints.

(a) Use the quotient rule to compute the first derivative:
$f'(x) = \frac{(2x)(x-2) - x^2(1)}{(x-2)^2} = \frac{x(x-4)}{(x-2)^2}$
The domain of $f$ excludes $x=2$, so even though $f'$ is undefined at $x=2$, it is not a critical point. $f'(x)=0$ at $x=0$ and $x=4$, both in the domain of $f$, so critical points are $x=0$ and $x=4$.
(b) The denominator $(x-2)^2$ is always positive for $x \neq 2$, so the sign of $f'$ matches the sign of $x(x-4)$. Testing intervals: $f'>0$ for $x<0$, $f'<0$ for $0<x<2$ and $2<x<4$, and $f'>0$ for $x>4$. At $x=0$, $f'$ changes from positive to negative → relative maximum. At $x=4$, $f'$ changes from negative to positive → relative minimum.
(c) On $[0,4]$, endpoints are $x=0$ and $x=4$, which we already classified, and $x=2$ is not in the domain. The relative extrema are: relative maximum at $(0, 0)$ and relative minimum at $(4, 8)$.

📐 Worked Example

A small business models its daily profit from selling $x$ units of a product as $P(x) = -0.001x^3 + 0.12x^2 + 3x - 80$, where $P(x)$ is measured in dollars, and $0 \leq x \leq 100$. Use the First Derivative Test to find the production level that gives a relative maximum daily profit, and interpret your result.

Compute the first derivative:
$P'(x) = -0.003x^2 + 0.24x + 3$
Find critical points: Set $P'(x)=0$, which simplifies to $x^2 - 80x - 1000 = 0$. Solving with the quadratic formula gives one positive critical point at $x \approx 91$, which is inside the interval $[0, 100]$.
Sign analysis: Test $x=10$: $P'(10) = 5.1 > 0$. Test $x=95$: $P'(95) \approx -1.28 < 0$.
Classification: $P'$ changes from positive to negative at $x \approx 91$, so this is a relative maximum. $P(91) \approx 482$ dollars.
Interpretation: Producing approximately 91 units per day gives the business a relative maximum daily profit of roughly \$482.

Common Pitfalls

Why: 学生自动将任何$f'$不存在的点标记为临界点，而不检查$f$的定义域

Why: 学生混淆了临界点的条件和极值的证明要求

Why: 学生学到临界点是内点，因此在分类相对极值时完全忽略端点

Why: 学生有时会不小心测试了临界点而非区间内的点，得到导数为零，错误得出符号不变的结论

Why: 学生在考试日快速解题时会混淆函数值和导数值

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一阶导数判别法判断相对极值 — AP 微积分 BC

1. 一阶导数判别法的核心概念 ★★☆☆☆ ⏱ 3 min

2. 内部临界点的符号分析 ★★☆☆☆ ⏱ 4 min

3. $f'$不存在处的临界点 ★★★☆☆ ⏱ 4 min

4. 端点极值的一阶导数判别法 ★★★☆☆ ⏱ 4 min

5. AP风格的额外例题 ★★★★☆ ⏱ 6 min

Common Pitfalls

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